Clausius-Mossotti relation

The Clausius-Mossotti relation connects the relative permittivity ε_r to the polarizability α of the molecules constituting a slab of dielectric.

Internal electric field in dielectric

Slab of dielectric (green) in external electric field E. Microscopic spherical cavity (size greatly exaggerated in drawing) of radius r in dielectric.

A macroscopic slab of dielectric is placed in an outer electric field E (the z-direction) that polarizes the dielectric, so that a surface charge density σ_p is created. Since E "pushes" positive charge and "pulls" negative charge, the sign of the charge density on the outer surface is as is indicated in the figure. The polarization vector P points by definition from negative to positive charge. The surface charge density σ_p is in absolute value equal to |P|.

An infinitesimally small spherical cavity of radius r is made in the dielectric and inside this cavity there is vacuum with permittivity (electric constant) ε₀. Because of electric neutrality the surface of the cavity has the charge density (in absolute value),

$\sigma (\theta )=\mathbf {n} \cdot \mathbf {P} =P\cos \theta ,$

where n is a unit length vector perpendicular to the surface of the cavity at the point θ. By definition, the positive direction of the normal n is outward.

An infinitesimal element $d\Omega =r^{2}\;\sin \theta \;d\theta d\phi$ on the surface of the cavity gives a contribution to the internal electric field E_int parallel to n,

$d\mathbf {E} _{\mathrm {int} }=\mathbf {n} {\frac {\sigma (\theta )d\Omega }{4\pi \epsilon _{0}r^{2}}}={\frac {P\cos \theta d\Omega }{4\pi \epsilon _{0}r^{2}}}={\frac {P\cos \theta \sin \theta d\theta d\phi }{4\pi \epsilon _{0}}}.$

The z-component of this contribution to the field is

$dE_{\mathrm {int} ,z}={\frac {P}{4\pi \epsilon _{0}}}(\cos \theta )^{2}\sin \theta d\theta d\phi .$

Integration over the whole surface gives

$E_{\mathrm {int} ,z}=\iint dE_{\mathrm {int} ,z}=-{\frac {P}{4\pi \epsilon _{0}}}\int _{0}^{\pi }(\cos \theta )^{2}d\cos \theta \int _{0}^{2\pi }d\phi ={\frac {1}{3\epsilon _{0}}}P.$

The total electric field in the cavity is in the z direction and has magnitude

$E_{\mathrm {tot} }=E+E_{\mathrm {int} }=E+{\frac {1}{3\epsilon _{0}}}P$